written 6.3 years ago by |
x, y = 2.5, 2.5
x1,y1 = 2, 2
x2,y2 = 4, 2
x3,y4 = 4, 3
x4,y4 = 2, 4
Converting into-local coordinate,
Now, ϕ1=14(1−ξ)(1−η)
ϕ2=14(1+ξ)(1−η)
ϕ3=14(1+ξ)(1−η)
ϕ4=14(1+ξ)(1−η)
Now, ˉx=∑ϕjˉxj
∴ ˉx = ϕ1ˉx1+ϕ1ˉx2+ϕ1ˉx3+ϕ1ˉx4
0.5 = 14(1−ξ)(1−η)ˉx1+14(1+ξ)(1−η)ˉx2+14(1+ξ)(1+η)ˉx3+14(1−ξ)(1+η)ˉx1
i.e. 12=14(1+ξ)(1−η)∗2+14(1−ξ)(1+η)∗2
i.e 12=12(1+ξ)(1−η)+12+(1+ξ)(1+η)
1=1+ξ+η−ηξ+1+ξ+η+ηξ
1=2+2ξ
ξ−12
= (1 + ξ)(1 − η) + (1 + ξ)(1 + η)
Now, ˉy=∑ϕjˉyj
0.5=140.5=(1−ξ)(1−η)ˉy1+14(1+ξ)(1−η)barx2+14(1+ξ)(1+η)ˉx3+14(1−ξ)(1+η)ˉx4
0.5 =14(1+ξ)(1+η)+14(1−ξ)(1+η)
12=14[1+ξ+η+ηξ+1−ξ+η−ηξ
∴ 2 = 2 +2η=>2η=0=>η=0
Therefore, point P is given by (-1/2,0) in ξ,−ηcoordinate,
To compute the result,
ϕ1=14(1−ξ)(1−η)=14∗(1+12)∗1=38
ϕ2=14(1−ξ)(1−η)=14∗(1+12)∗1=18
ϕ3=14(1−ξ)(1−η)=14∗(1+12)∗1=18
ϕ4=14(1−ξ)(1−η)=14∗(1+12)∗1=38
Now,
ϕ1+ϕ2+ϕ3+ϕ4=38+18+18+38=88
ϕ1+ϕ2+ϕ3+ϕ4
Now, Temperature at point P,
TP=ϕ1T1+ϕ2T2+ϕ3T3+ϕ4T4
∴ T1=100∘CT2=60∘CT3=50∘CT4=90∘C
∴ TP=38∗100+18∗60+18∗50+38∗90
TP=85∘C